Can anyone solve this with software?

1. Jul 18, 2011

wil3

I have tried mathematica, and it says it lacks the means to solve it:

The function
$$g6_{\mu,\sigma}[x]$$
represents the SIXTH derivative of a normal distribution with unspecified parameters. I am looking to solve the relation:
$$g6_{\mu,\sigma}[\mu+ \frac{\delta}{2}] + g6_{\mu,\sigma}[\mu - \frac{\delta}{2}] = 0$$
in terms of $$\delta$$. I have a feeling that the answer does not depend on mu, just sigma.

The application is finding the minimum separation required between the central peaks of two 4-derivative gaussian curves such that there occur no inflections on the consolidated central peak. This is related to Sparrow's criterion.

Thank you very much for any help.

2. Jul 18, 2011

Bill Simpson

Check all this carefully to make certain I have not misunderstood or made a mistake.

In[1]:= g[x_]:=E^-((x-mu)^2/(2 sig^2));
In[2]:= d = Derivative[6][g];
In[3]:= Simplify[d[mu + del/2] + d[mu - del/2]]

Notice mu does not appear in the result. Perhaps that settles your question.

If not then notice only the 6th order polynomial can supply solutions and del always appears to even powers so you can use tricks to solve a cubic and then find square roots for the six original solutions. But try brute force.

In[4]:= Simplify[Solve[del^6-60*del^4*sig^2+720*del^2*sig^4-960*sig^6 == 0, del]]

Now you have to decide which of those six apply to your problem.

3. Jul 18, 2011

wil3

Yeah, another method I tried gets those imaginary solutions as well. My only issue with that is that the graph very definitely has a real solution, since certain values of sigma cause the graph to hit the x-axis and then cross it. I tried running a bunch of points to approximate it, and I got a number somewhere around 1.33, but I would like to know the full form.

Can you explicate the cubic solution approach?

4. Jul 18, 2011

Bill Simpson

Since all your del are to even powers you effectively substitute newdel=del^2 and get

Simplify[Solve[newdel^3-60*newdel^2*sig^2+720*newdel*sig^4-960*sig^6 == 0, newdel]]

Then you get three nice and often somewhat more compact solutions for newdel. Each of those is the square of a solution for del and there are positive and negative square roots. For some problems you can determine by inspection which newdel solution applies.

In some cases complex values in a solution will wash out to give you real solutions when you put in your actual parameters.

If I take the first of those 6 solutions, substitute sig->2, FullSimplify, take the resulting polynomial and plot it I see 6 nice real solutions. So perhaps the complex values are just used in the presentation of the solutions and aren't giving you complex roots after all.

5. Jul 18, 2011

wil3

Okay, so I just used the Reduce function and got \[Delta] == -\[Sigma] Sqrt[Root[-960 + 720 #1 - 60 #1^2 + #1^3 &, 1]]

I do not know how to interpret this output... once we get to the Slot operator, my understanding begins to run out. However, this expression evaluates numerically to 1.22`sigma, which is the correct answer.

Thanks for your help. You're a pro.